P1576 minimum cost

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This question is graph theory shortest path, but we have to change it in some detail

For example, because it is considered the exchange rate, we initialize will be 0

We also altered relaxation operation

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Also, subject to the exchange rate (and shaping)

So we wanted to change decimals

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And, the exchange rate is in our operation "loss"

So, we are left with part of the exchange rate is 1- (readily available)

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DOUBLE warning

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These are special attention

In fact, the range of data as well as an array of pits only 100,000 RE

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Topic links : Miku

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#include<iostream>
#include<queue>
#include<stack>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;
struct bian {
    int to;
    int next;
    double v; //注意，是double
} b[1000010];
double dis[1000010];//double警告
queue<int> q;
int head[1000001];
int pb;
void add(int form,int to,double v) {//double警告
    pb++;
    b[pb].to=to;
    b[pb].v=1-v;//省很多事的
    b[pb].next=head[form];
    head[form]=pb;
}
bool vis[1000001];
int main() {
    memset(dis,0,sizeof(dis));
    cin>>n>>m;
    int x,y,z;
    for(int i=1; I <= m; ++ I) { 
        CIN >> >> X >> Y Z; 
        the Add (X, Y, ( Double ) Z / 100 ); 
        the Add (Y, X, ( Double ) Z / 100 ); // must be processed into a decimal 
    } 
    CIN >> >> X Y; 
    { 
        // the SPFA portion 
        DIS [X] = . 1 ; 
        VIS [X] = . 1 ; 
        q.push (X); // initialize 
        {
             the while (Q. size ()) {
                 int U = q.front (); 
                q.pop (); 
                VIS [U] =0;
                for(int i=head[u]; i; i=b[i].next) {
                    int v=b[i].to;
                    if(dis[v]<dis[u]*b[i].v) {//特殊操作
                        dis[v]=dis[u]*b[i].v;//特殊操作
                        {
                            if(!vis[v]) {
                                q.push(v);
                                vis[v]=1;
                            }
                        }
                    }
                } 

            }
        }
    } 
    Double ANS = 100 / DIS [Y]; // Double warning 
    the printf ( " % .8lf " , ANS); // decimal Oh 
    return  0 ; 
}

AC

 

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That's all.